Byju's Answer

Standard X

Mathematics

Distance between Two Points

Origin is shi...

Question

Origin is shifted to the point P without changing the directions of the axes. If the coordinates of Q with respect to the old axes, new axes are $(2, - 1, 4)$ and $(3, 1, 2)$ respectively, then $P_{x} + P_{y} + P_{z} =$ ?

Open in App

Solution

$\begin{matrix} 2 = 3 + h \Rightarrow h = - 1 - 1 + 1 + k \Rightarrow k = - 2 4 = 2 + β \Rightarrow β = 2 P_{x} = - 1 p_{y} = - 2 P_{z} = 2 (h, k, β) \equiv (p_{x}, p_{y}, p_{z}) p_{x} + p_{y} + p_{z} = - 1 \end{matrix}$

Suggest Corrections

Similar questions

Q. If the origin is shifted to

(1, 2, - 3)

without changing the direction of the axes, then find the new coordinates of the point

(0, 4, 5)

with respect to new frame.

Q. If the coordinates of P are

(2, - 1, 4)

. The origin is shifted to

(1, 1, 1)

keeping the direction of the axes same. The new coordinates are

If origin is shifted to the point $(2, 3)$ without rotation of axes then the coordinates of the point $P$ which divides the join of $A (4, 8)$ and $B (7, 14)$ in the ratio $1 : 2$ or $2 : 1$ with respect to the new system of coordinates can be

Q. Assertion :

A B C D

is a square with vertices

(0, 0), (1, 0), (1, 1)

and

(0, 1) . P, Q, R

and

S

are the points which divide

A B, B C, C D, D A

respectively in the ratio

2 : 1

. If the origin is shifted to the centre of the square

A B C D

without rotation of axes, area of the square

P Q R S

in the new system of coordinates is

\frac{5}{9}

square units. Reason: If the origin is shifted to the point

P (α, β)

without rotation of axes then the distance between any two given points remains unchanged in the new system of coordinates

Q. Keeping the origin constant axes are rotated at an angle

30^{o}

in negative direction, if then new coordinates of the point are

(2, 1)

then its coordinate are with respect to old axis are

Origin is shifted to the point P without changing the directions of the axes. If the coordinates of Q with respect to the old axes, new axes are (2,−1,4) and (3,1,2) respectively, then Px+Py+Pz=?

2=3+h⇒h=−1−1+1+k⇒k=−24=2+β⇒β=2Px=−1py=−2Pz=2(h,k,β)≡(px,py,pz)px+py+pz=−1

Origin is shifted to the point P without changing the directions of the axes. If the coordinates of Q with respect to the old axes, new axes are $(2, - 1, 4)$ and $(3, 1, 2)$ respectively, then $P_{x} + P_{y} + P_{z} =$ ?

$\begin{matrix} 2 = 3 + h \Rightarrow h = - 1 - 1 + 1 + k \Rightarrow k = - 2 4 = 2 + β \Rightarrow β = 2 P_{x} = - 1 p_{y} = - 2 P_{z} = 2 (h, k, β) \equiv (p_{x}, p_{y}, p_{z}) p_{x} + p_{y} + p_{z} = - 1 \end{matrix}$